The plane mirror is fixed at a height A above the bottom of a beaker containing water (v) upto a height d. The position of image formed formed from the mirror is
A) \[h+\frac{d}{\mu }\]
B) \[\frac{h+d}{\mu }\]
C) \[\frac{h-2d}{\mu }\]
D) \[h-\frac{d}{\mu }\]
E) \[h-d+\frac{d}{\mu }\]
Correct Answer: E
Solution :
The bottom of beaker appears to be shifted up by distance \[\Delta t=\left( 1-\frac{1}{\mu } \right)d\] The apparent distance of the bottom from the mirror is \[h-\Delta t=h-\left[ 1-\frac{1}{\mu } \right]d=h-d+\frac{d}{\mu }.\] The image is formed behind the mirror at a distance \[h-d+\frac{d}{\mu }.\]
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